Evaluate the following integrals:

We have,

$x^{2}+6 x+13=x^{2}+6 x+3^{2}-3^{2}+13$

$=(x+3)^{2}+4$

Sol, $\int \frac{1}{x^{2}+6 x+13} d x=\int \frac{1}{(x+3)^{2}+2^{2}} d x$

Let $x+3=t$

Then $d x=d t$

$\int \frac{1}{(\mathrm{t})^{2}+2^{2}} \mathrm{dt}=\frac{1}{2} \tan ^{-1} \frac{\mathrm{t}}{2}+\mathrm{c}$

[since, $\left.\int \frac{1}{x^{2}+(a)^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right]$

$\frac{1}{2} \tan ^{-1} \frac{x+3}{2}+c$